Functions 2 Question 7
8. Let $f(x)=\frac{\alpha x}{x+1}, x \neq-1$. Then, for what value of $\alpha$ is $f[f(x)]=x ?$
(a) $\sqrt{2}$
(b) $-\sqrt{2}$
(c) 1
(d) -1 $-1, \quad x<0$
(2001, 1M)
Show Answer
Answer:
Correct Answer: 8. (d)
Solution:
- Given, $f(x)=\frac{\alpha x}{x+1}$
$$ \begin{array}{rlrl} & f[f(x)]=f \frac{\alpha x}{x+1}=\frac{\alpha \frac{\alpha x}{x+1}}{\frac{\alpha x}{x+1}+1} \\ & =\frac{\frac{\alpha^{2} x}{x+1}}{\frac{\alpha x+(x+1)}{x+1}}=\frac{\alpha^{2} x}{(\alpha+1) x+1}=x \text { [given] …(i) } \\ \Rightarrow \quad & \alpha^{2} x=(\alpha+1) x^{2}+x \\ \Rightarrow \quad & x\left[\alpha^{2}-(\alpha+1) x-1\right]=0 \\ \Rightarrow \quad & x(\alpha+1)(\alpha-1-x) & =0 \\ \Rightarrow \quad & & \text { and } \quad \alpha+1 & =0 \\ \Rightarrow & & \alpha & =-1 \end{array} $$
But $\alpha=1$ does not satisfy the Eq. (i).
